Question

determine if the given function has any points of discontinuity. explain your reasoning.

(f(x) = \frac{x^2 - b^2}{x - b})

there is a point of discontinuity at (x = b) because the denominator has the factor (x - b).
there are points of discontinuity at both (x = -b) and (x = b) because the numerator has factors of (x + b) and (x - b).
there is a point of discontinuity at (x = -b) only because the factor of (x - b) is common to both the numerator and denominator.
there is a point of discontinuity at (x = b) only because the factor of (x - b) is common to both the numerator and denominator and factors out.

Explanation:

Identify domain restrictions from the denominator

$$ x - b = 0 \implies x = b $$

Factor the numerator to analyze the discontinuity

$$ f(x) = \frac{(x - b)(x + b)}{x - b} $$

Determine the type of discontinuity

$$ \text{Since } (x - b) \text{ is common to both numerator and denominator, it factors out, creating a removable discontinuity at } x = b. $$

Answer:

• There is a point of discontinuity at \(x = b\) because the denominator has the factor \(x - b\).
• There are points of discontinuity at both \(x = -b\) and \(x = b\) because the numerator has factors of \(x + b\) and \(x - b\).
• There is a point of discontinuity at \(x = -b\) only because the factor of \(x - b\) is common to both the numerator and denominator.
• There is a point of discontinuity at \(x = b\) only because the factor of \(x - b\) is common to both the numerator and denominator and factors out. (Correct answer)